Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{-z^2 - 5z}{-3z - 3} \div \dfrac{3z^3 + 45z^2 + 150z}{3z^2 + 48z + 180} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{-z^2 - 5z}{-3z - 3} \times \dfrac{3z^2 + 48z + 180}{3z^3 + 45z^2 + 150z} $ First factor out any common factors. $n = \dfrac{-z(z + 5)}{-3(z + 1)} \times \dfrac{3(z^2 + 16z + 60)}{3z(z^2 + 15z + 50)} $ Then factor the quadratic expressions. $n = \dfrac {-z(z + 5)} {-3(z + 1)} \times \dfrac {3(z + 10)(z + 6)} {3z(z + 10)(z + 5)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-z(z + 5) \times 3(z + 10)(z + 6) } {-3(z + 1) \times 3z(z + 10)(z + 5) } $ $n = \dfrac {-3z(z + 10)(z + 6)(z + 5)} {-9z(z + 10)(z + 5)(z + 1)} $ Notice that $(z + 10)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-3z\cancel{(z + 10)}(z + 6)(z + 5)} {-9z\cancel{(z + 10)}(z + 5)(z + 1)} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $n = \dfrac {-3z\cancel{(z + 10)}(z + 6)\cancel{(z + 5)}} {-9z\cancel{(z + 10)}\cancel{(z + 5)}(z + 1)} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $n = \dfrac {-3z(z + 6)} {-9z(z + 1)} $ $ n = \dfrac{z + 6}{3(z + 1)}; z \neq -10; z \neq -5 $